Wouldn't a heap be ideal for this ?
On Mon, Jul 11, 2011 at 3:35 PM, John Reid j.r...@mail.cryst.bbk.ac.ukwrote:
I have a procedure that generates N x M values sequentially. I want to
store the N largest ones and discard the others. Obviously I can store all
the values in a vector, sort it
You can use priority_queue in STL. The size needs to be limited to N
elements. At any point the the N elements in the heap will give the largest
N
elements processed so far.
On Mon, Jul 11, 2011 at 4:41 PM, John Reid j.r...@mail.cryst.bbk.ac.ukwrote:
On 11/07/11 12:07, abhijith reddy wrote
p[3] = 'E'
p[1] = 'A'
p[3]-p[1] = 4
?
On Sat, Jul 2, 2011 at 7:10 PM, KK kunalkapadi...@gmail.com wrote:
10. What does the following fragment of C-program print?
char c[] = GATE2011;
char *p =c;
printf(%s, p + p[3] - p[1]);
(A) GATE2011 (B) E2011 (C) 2011 (D) 011
Answer: - (C)
why is
+1
On Fri, Jun 24, 2011 at 1:06 PM, rizwan hudda rizwanhu...@gmail.com wrote:
Add Donald Knuth, Rajeev Motwani, Manindra agrawal, Richard Karp, Vijay V
Vazirani to the list.
Peter is no doubt a great algorithm champion, and so is ACRush. But they
are definetely not the
pioneers of the
while(1):
try:
# code
#
#
except EOFError: break
On Wed, May 25, 2011 at 8:41 PM, Vishnutej
mylavarapu.vishnu...@gmail.comwrote:
Hello everyone,
I'm new to python.How can I check the EOF for inputs in SPOJ?
Thanks in advance!!
-Vishnutej
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, 2011 at 8:42 PM, abhijith reddy
abhijith200...@gmail.comwrote:
while(1):
try:
# code
#
#
except EOFError: break
On Wed, May 25, 2011 at 8:41 PM, Vishnutej
mylavarapu.vishnu...@gmail.com wrote:
Hello everyone,
I'm new to python.How can I check the EOF for inputs
Let *E(n)* be the *expected* number of hits needed sort *'n'*
*misplaced*numbers.
The optimal strategy is to hold the numbers that are already in place and
shuffle the remaining.
Now after a shuffle assume that x numbers are still out of place.
Hence we get the following recurrence.
E(n) = Sum[
O(N^3) DP
-
char str[MAX];
int dp1[MAX][MAX],dp2[MAX][MAX];
int isPalin(int low,int high)
{
if(low=high) return 1;
if(dp1[low][high]!=-1) return dp1[low][high];
return
possible
positions between i j and take minimum of all.
On Fri, May 6, 2011 at 8:23 PM, sourabh jakhar sourabhjak...@gmail.comwrote:
can you explain the logic
On Fri, May 6, 2011 at 8:02 PM, abhijith reddy
abhijith200...@gmail.comwrote:
O(N^3) DP
You could read input character by character using getchar_unlocked() untill
you hit a space or new line or EOF.
Alternatively you read the whole input at once using fread_unlocked() and
then process it as per your need.
On Wed, Apr 20, 2011 at 7:41 AM, shubham shubh2...@gmail.com wrote:
Hello
cont int MAX = 10005;
bool isPrime[MAX];
void sieve()
{
int lim=sqrt(MAX);
/* Initially Mark all numbers as Prime */
for(int i=2;iMAX;i++)
isPrime[i]=1;
for(int i=2;i=lim;i++)
if(isPrime[i])/* for each Prime */
for(int j=2*i;jMAX;j+=i)/* Cross
17^13 is odd
13*17 is odd
so isn't the answer simply 2 ?
On Thu, Mar 17, 2011 at 8:25 PM, saurabh singh saurab...@gmail.com wrote:
Find the smallest divisor of 17^13+13*17...
PS:please dont say 1
--
Saurabh Singh
B.Tech (Computer Science)
MNNIT ALLAHABAD
--
Saurabh Singh
copy data from the next node and then delete that next node.
Say you need to delete 3
1 - 2 - 3 - 4 - 5
1 - 2 - 4 - 4 - 5
*
Now delete node which is next to '*'.
On Sun, Mar 13, 2011 at 10:23 PM, UMESH KUMAR kumar.umesh...@gmail.comwrote:
hi
Given a singly Link list but
algorithm isn't required for C++ 4.0.*. The same code will give Compile
Error in C++ 4.3 and above
On Tue, Mar 8, 2011 at 10:42 PM, Logic King crazy.logic.k...@gmail.comwrote:
Well tried,
i have got the correct answer after working on it for almost 2 hours
here is my code
http://zobayer.blogspot.com/2010/03/small-bigint-library.html
On Fri, Feb 4, 2011 at 10:24 PM, Rahul Verma rahulverma@gmail.comwrote:
@jeeva
Can you pls explain me in detail or some link to the tutorial of
string manipulation in C++.
I googled about it but most of the links are in Python
simple dp
void solve(int *arr,int sz)
{
int ans[sz];
ans[sz-1]=-1;
for(int i=sz-2;i=0;i--)
{
if(arr[i]arr[i+1]) ans[i]=arr[i+1];
else ans[i]=ans[i+1];
}
for(int i=0;isz;i++)
printf(%d ,ans[i]);
}
On Sun, Jan 30, 2011 at 10:00 PM, ritu
I remember solving this @ spoj Here is an O(1) solution
#!/usr/bin/python
def solve(n):
val=1
for i in range(1,9):
val*=(n+i)
return float((n+9)/9.0-(40320.0/val))
cases=int(raw_input())
while(cases):
cases-=1
n=int(raw_input())
O(2^n)
On Sat, Jan 22, 2011 at 8:58 PM, Decipher ankurseth...@gmail.com wrote:
fun(n)
{
if(n=2)
return (1);
else
return ((fun(n-1)*fun(n-2));
}
find the order of complexity .
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Below is code for modular exponentation in general
// (a^b)%c
int modexp(int a,int b,int c)
{
int ans=1;
while(b)
{
if(b1) ans=(ans*a)%c;
a=(a*a)%c;
b=1;
}
return ans;
}
On Fri, Jan 21, 2011 at 3:27 PM, juver++ avpostni...@gmail.com wrote:
int l = 0, r = ...;
not to use log or power. isnt there any
approach using bitwise operators
On Fri, Jan 21, 2011 at 5:24 PM, Manmeet Singh mans.aus...@gmail.comwrote:
this will be O(log(n) * log(n)) solution
On Fri, Jan 21, 2011 at 4:29 PM, abhijith reddy abhijith200...@gmail.com
wrote:
Below is code
@snehal .. misread it .. my apologies.
On Fri, Jan 21, 2011 at 5:56 PM, abhijith reddy abhijith200...@gmail.comwrote:
O(lg(n)*lg(n)) is the complexity of the solution ! Not the solution.
M=3^x
We binary search on x and then compute 3^x in log(x) time using
exponentiation. Hence
I think its correct.
On Jan 19, 9:35 pm, nishaanth nishaant...@gmail.com wrote:
How about the following dynamic programming solution.
Let dp[i] be the max no of As with i keystrokes.
dp[i]=max(dp[i-1]+1,2*dp[i-3])
dp[N] is the required solution.
Correct me if i am wrong.
On Wed, Jan
binary search on n
On Fri, Apr 30, 2010 at 10:15 PM, Amit Agarwal lifea...@gmail.com wrote:
how do I compute n from this equation.
n 8lg(n)
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I guess she was asking that the per query complexity should be better that
O(n).
If that is the case then you can use any of these
Simple RMQ O(sqrt(n))
Segment/Interval Trees O(lgn)
Binary Indexed Trees O(lgn)
On Wed, Mar 31, 2010 at 12:58 PM, Rohit Saraf
rohit.kumar.sa...@gmail.comwrote:
First learn a language of your choice and then you can start solving over
there
On Wed, Mar 31, 2010 at 12:05 AM, scanfile rahul08k...@gmail.com wrote:
I am new to the world of programming. I have to solve the problems on
the spoj.pl , but I have no idea that how I would implement the
@All: Doing a sorting for O(n logn) and then binary search for x1 and x2 in
O(logn) will be less efficient than the simple solution of O(n). Think on
the data structure that can optimize it.
Is it possible in time complexity O(n)?
If you want to do the operation just once then it cannot
You can use a TRIE .. Structure can be something like this
struct trie
{
int count; // no of occurences
char *child[SIZE];
};
when ever u insert ( it will take just O(length) time) .. just increment
count by 1
For each query (also O(length) time) the no of occurrences of the word will
Yes i have .. and i have the worst time in the rank page :)
On Fri, Jan 29, 2010 at 8:51 PM, fundoonick fundoon...@yahoo.co.in wrote:
I tried this problem using the method of solving using eulerphi values.
I calculate values of eulerphi in O(nlogn) and then use them to calculate
gcdsum using
Hello
Let gcdsum[n] = gcd(1,1)+gcd(1,2)+gcd(1,3)+ ... +gcd(n,n)
Also gcdsum[n] can be evaluated using :
gcdsum[n] = n*sum(eulerphi(d)/d) for all d | n
Now to compute all gcdsum values upto n we first need to precompute
all eulerphi and then use a sieve like algorithm to make a table of
all
*TRIE*: Using a trie we would get a linear time solution but i think the
memory requirement would be huge ..
On Mon, Dec 14, 2009 at 11:02 PM, vicky mehta...@gmail.com wrote:
Given a file with a lot of words (10 million) find out the top 10%
most
frequently occuring words.
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You
I heard that there's a sieve algorithm whose complexity is O(n).
Can any give me the pseudo code for that ?
Thank u !
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To
+ 10 + 1 = 16
On Wed, Nov 4, 2009 at 11:23 AM, abhijith reddy abhijith200...@gmail.com
wrote:
Is there a way to find the sum of the Kth series ( Given below)
K=0 S={1,2,3,4,5,6,}
K=1 S={1,2,4,7,11,16..} common diff = 1,2,3,4 5 ...
K=2 S={1,2,4,8,15,26...} common diff = 1,2,4,7 11
Is there a way to find the sum of the Kth series ( Given below)
K=0 S={1,2,3,4,5,6,}
K=1 S={1,2,4,7,11,16..} common diff = 1,2,3,4 5 ...
K=2 S={1,2,4,8,15,26...} common diff = 1,2,4,7 11... (series with
K=1)
K=3 S={1,2,4,8,16,31...} common diff = 1,2,4,8 15... (series with
K=2)
Note
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